setting the dropping resistor in fixed bias -- ?

[jimipage]

OK, so now that some things have been clarified, I'm still confused as to how to actually calculate and approximate not only the dropping resistor, but the entire circuit. It seems like skyboltone was on the right path. Can anybody maybe give a step-by-step on how to do this? I really need the help.

thanks

[Jana]

step by step:

Use these values: 180k if you are sure the customer won't want to mod it in the future to stick in the super dooper van halen reissue el34 tubes from fleabay. If it is a possibility, use 220k and jump with 1 meg, then you can just tell them, "Clip the 1 meg."

two caps, 10uf each with the 15k between them. Don't go higher than 10uf, no advantage in filtering and the time constant gets way to long with the high value resistor.

If you're going to really have it fixed, as in no trimmer, use a 33K in series with a 10 or 25k pot, adjust the pot when you have the amp fired up to get the bias you want. Remove pot and measure it to find the fixed value you need.

If you are going to have a trimmer, use a 10 or 25K trimmer along with the 33K. This should get you real close with the trimmer close to the middle, if its not, adjust the 33K to get you there.

After building close to a hundred amps, most all of them with this bias arrangement, this is what works for me. I use this even when I have the dedicated bias tap and don't have to do a full wave bridge.

[jimipage]

Thanks, Jana, for your replies. I appreciate the help. I will, in fact, employ the circuit you describe. But (Please don't take offense at this), I was asking how to calculate a fixed-bias circuit. Basically, how/why was a 220k/180k/etc. dropping resistor chosen, as well as the 15k between the filter caps, 33k in series with a pot, etc....? I want to know the 'why and how' to get those numbers so that I can approach fixed-bias with other tubes as well.

[skyboltone]

Skyboltone,

We multiply peak voltages by 0.707 to obtain RMS voltages.

Since the secondary voltages of power transformers is already given in V RMS, why are you multiplying it by 0.707?

This transformer has secondary RMS voltages of 350-0-350, when the primary is 115 VAC. It will have a Vpeak of about 495, probably higher as our line voltages these days are 120VAC and above.

VAC is the same as Vrms and not the same as Vpeak. 120VAC = 170Vpeak.

Duncan Amps has a power supply designer that people might find useful for this application:

http://www.duncanamps.com/psud2/index.html

DC output voltage through half wave capacitor input filter equals .707 times RMS input voltage. Full wave = 1.41 +/-